Laws or motions and a truck that won't move.

  • Thread starter Thread starter azertyq
  • Start date Start date
  • Tags Tags
    Laws Truck
Click For Summary
SUMMARY

The discussion revolves around a physics problem involving a tow truck exerting a force of 2500N on a car, with the goal of determining the forces in the towing crane's members while both vehicles remain stationary. The user initially calculated the forces Fa and Fb incorrectly, obtaining values of 3214N and 1632N, respectively, while the textbook states the correct values are Fa=3.83kN and Fb=3.37kN. The conversation highlights the importance of double-checking calculations and understanding the principles of equilibrium and Newton's First Law of Motion.

PREREQUISITES
  • Understanding of Newton's First Law of Motion
  • Knowledge of equilibrium in physics
  • Ability to perform vector calculations
  • Familiarity with free body diagrams (FBD)
NEXT STEPS
  • Review vector resolution techniques in physics
  • Study equilibrium conditions in static systems
  • Learn about free body diagram analysis
  • Explore common mistakes in force calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, engineers involved in statics, and anyone interested in understanding forces and equilibrium in real-world applications.

azertyq
Messages
2
Reaction score
0
I have been bothered with a problem that I know should not be monopolizing as much time as it has. It just annoys me how I keep missing the little thing that would make the problem more understandable.

Anyways, here is the problem, with my attempt of a solution:

a tow truck pulls on a car with a 2500N force directed along the cable used for towing. The purpose of the problem is to find the force in the members of the 'towing crane' used, assuming that both the car and the truck are not moving. (the attached FBD should make more sense.. sorry)

Here are the calculations and results I got:

(sum of forces F) F = ma = 0 // at the point C where the three forces are joined

--> Fx= 0 and Fy=0


F1= -2500* cos(40) i - 2500*sin(40) j // finding the vector corresponding to each force

Fa= - Fa j

Fb= Fb*cos(50) i - Fb*sin(50) j

Fx= [ (-2500* cos(30) + Fb*cos(50)]= 0 //applying the equilibrium equation
--> Fb = 2500* cos(30) / *cos(50)
Fb= 1632 N
Fy= [ - 2500*sin(30) - Fa - Fb*sin(50)] = 0

--> Fa = -(2500*sin(30) + Fb*sin(50))
Fa = -(2500*sin(30) + 2979*sin(50))
Fa= 3214 N


It seems logic to me that it'd be the correct resolution path, but according to the textbook, the correct values for Fa and Fb are

Fa= 3.83kN and Fb=3.37kN


Quite different from what I got indeed.

If anyone could help me with this problem, it'd be great. I tried. I really did. :(
 

Attachments

  • physics.JPG
    physics.JPG
    14.4 KB · Views: 536
Last edited:
Physics news on Phys.org
I feel kind of stupid after finding out that I had jut made a computation error.
I guess things like that are bound to happen. >_<
lol

Sorry to have bothered you guys.
 



Thank you for sharing your problem and your attempted solution. It can be frustrating when we put in effort to solve a problem and still come up with incorrect results. In this case, it seems like there may be a mistake in your calculations or assumptions. It is always a good idea to double check your work and make sure you are using the correct formulas and units.

In terms of laws or motions, there are several principles that can help us understand why the truck is not moving in this scenario. One is Newton's First Law of Motion, which states that an object at rest will remain at rest unless acted upon by an external force. In this case, the truck is not moving because the forces acting on it are balanced, resulting in a net force of zero.

Another relevant concept is equilibrium, which is when all forces acting on an object are balanced and there is no net force. In your solution, you correctly identified that the sum of forces in the x and y directions should be equal to zero for equilibrium to be achieved. However, it is possible that there was a mistake in your calculations that led to the incorrect values for Fa and Fb.

In the end, it is important to remember that mistakes and challenges are a normal part of problem-solving. It may be helpful to go back and review the problem, check your calculations, and seek help from a teacher or classmate if needed. Keep persevering and you will eventually find the solution.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 9 ·
Replies
9
Views
14K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 8 ·
Replies
8
Views
9K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
24
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
17
Views
3K